Formula for diarsenic pentoxide
As2O5
Answer:
The answer is 18.12KJ is required to vaporise 48.7 g of dichloromethane at its boiling point
Explanation:
To solve the above question we have the given variable as follows
ΔHvap = heat of vaporisation of dichloromethane per mole = 31.6KJ/mole
However since the heat of vaporisation is the heat to vaporise one mole of dichloromethane, then, for 48.7 grams of dichloromethane, we have.
The number of moles of dichloromethane present = 48.7/84.93 = 0.573 moles
Therefore, the amount of heat required to vaporise 48.7 grams of dichloromethane at its boiling point is 31.6KJ/mole×0.573moles =18.12KJ
Answer: 75%
Explanation:
The following information can be gotten from the question:
Waste = 70kg
Theoretical yield = 280kg
Therefore, the actual yield will be the difference between the theoretical yield and the waste which will be:
= 280kg - 70kg = 210kg
The percent yield will now be:
= Actual yield / Theoretical yield × 100
= 210/280 × 100
= 3/4 × 100
= 75%
Answer:
The correct answer is 0.3582 M.
Explanation:
Based on the given information, the volume of KOH given is 19.55 ml, the volume of sulfuric acid given is 20.20 ml, and the molarity of sulfuric acid is 0.3467 M. There is a need to find the molarity of KOH.
The formula to use in the given case is,
M1V1 = M2V2
Here M1 is the molarity of KOH, V1 is the volume of KOH, M2 is the molarity of sulfuric acid, and V2 is the volume of H2SO4.
Here V1 is 19.55 ml, M2 is 0.3467 M, and V2 is 2020 ml.
Now putting the values in the equation we get,
M1 * 19.55 = 0.3467 * 20.20
M1 * 19.55 = 7.00334
M1 = 7.00334/19.55
M1 = 0.3582 M
Hence, the molarity of the given KOH is 0.3582.
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hope this helps<span />
